The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.2$ years; the standard deviation is $0.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living longer than $1.1$ years.
Solution: $3.2$ $2.5$ $3.9$ $1.8$ $4.6$ $1.1$ $5.3$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $3.2$ years. We know the standard deviation is $0.7$ years, so one standard deviation below the mean is $2.5$ years and one standard deviation above the mean is $3.9$ years. Two standard deviations below the mean is $1.8$ years and two standard deviations above the mean is $4.6$ years. Three standard deviations below the mean is $1.1$ years and three standard deviations above the mean is $5.3$ years. We are interested in the probability of a lizard living longer than $1.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lizards will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $1.1$ years and the other half $({0.15\%})$ will live longer than $5.3$ years. The probability of a particular lizard living longer than $1.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.